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<h3 class="heading"><span class="type">Paragraph</span></h3>
<p>Introduce a change of variables:</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
{\bf x}={\bf T}\, \vec{y}
\end{equation*}
</div>
<p class="continuation">where <span class="process-math">\(\vec{y}\)</span> are new variables. Then <span class="process-math">\({\bf x}^{\prime}={\bf A}\,{\bf x}\)</span> becomes</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
{\bf T}\, \vec{y}^{\prime}={\bf A}\, {\bf T}\,\vec{y}.
\end{equation*}
</div>
<p class="continuation">Multiplying the inverse of <span class="process-math">\({\bf T}\)</span> on both sides:</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\vec{y}^{\prime}={\bf T}^{-1} {\bf A} \,{\bf T}\,\vec{y}.
\end{equation*}
</div>
<p class="continuation">According to what we have learnt in linear algebra,</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
{\bf T}^{-1} {\bf A} {\bf T}=\left(
\begin{array}{cccc}
r_1 &amp; 0 &amp; \cdots &amp; 0\\
0 &amp; r_2 &amp; \cdots &amp; 0\\
\vdots &amp; \vdots &amp; \ddots &amp; \vdots\\
0 &amp; 0 &amp; \cdots &amp; r_n
\end{array}
\right).
\end{equation*}
</div>
<p class="continuation">Thus,</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\vec{y}=\left(
\begin{array}{c}
C_1 e^{r_1 t}\\
C_2 e^{r_2 t}\\
\vdots\\
C_n e^{r_n t}\\
\end{array}
\right),
\end{equation*}
</div>
<p class="continuation">and</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
{\bf x}={\bf T} \, \vec{y}={\bf T} \,\left(
\begin{array}{c}
C_1 e^{r_1 t}\\
C_2 e^{r_2 t}\\
\vdots\\
C_n e^{r_n t}\\
\end{array}
\right).
\end{equation*}
</div>
<span class="incontext"><a href="sec6_5.html#p-282" class="internal">in-context</a></span>
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